Integration 211
Example 3
If
dy
dx
= 3 x^2 +2, and the point (0,−1) lies on the graph ofy, findy.
Since
dy
dx
= 3 x^2 +2, thenyis an antiderivative of
dy
dx
. Thus,
y=
∫ (
3 x^2 + 2
)
dx=x^3 + 2 x+C. The point (0,−1) is on the graph ofy.
Thus,y =x^3 + 2 x+C becomes− 1 = 03 +2(0)+C orC =−1. Therefore, y=x^3 +
2 x−1.
Example 4
Evaluate
∫ (
1 −
1
√ (^3) x 4
)
dx.
Rewrite as
∫ (
1 −
1
x^4 /^3
)
dx=
∫ (
1 −x−^4 /^3
)
dx
=x−
x−^1 /^3
− 1 / 3
+C=x+
3
√ (^3) x+C.
Example 5
Evaluate
∫
3 x^2 +x− 1
x^2
dx.
Rewrite as
∫ (
3 +
1
x
−
1
x^2
)
dx=
∫ (
3 +
1
x
−x−^2
)
dx
= 3 x+ln|x|−
x−^1
− 1
+C= 3 x+ln|x|+
1
x
+C.
Example 6
Evaluate
∫ √
x
(
x^2 − 3
)
dx.
Rewrite
∫
x^1 /^2
(
x^2 − 3
)
dx=
∫ (
x^5 /^2 − 3 x^1 /^2
)
dx
=
x^7 /^2
7 / 2
−
3 x^3 /^2
3 / 2
+C=
2
7
x^7 /^2 − 2
√
x^3 +C.
Example 7
Evaluate
∫ (
x^3 −4 sinx
)
dx.
∫ (
x^3 −4 sinx
)
dx=
x^4
4
+4 cosx+C.
Example 8
Evaluate
∫
(4 cosx −cotx)dx.
∫
(4 cosx−cotx)dx=4 sinx−ln
∣∣
sinx
∣∣
+C.